In a radioactive material, fraction of active material remaining after time t is frac{9}{16} fractio

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13.Nuclei

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In a radioactive material, fraction of active material remaining after time $t$ is $\frac{9}{16}$ The fraction that was remaining after $\frac{t}{2}$ is

A

$\frac{3}{4}$

B

$\frac{7}{8}$

C

$\frac{4}{5}$

D

$\frac{3}{5}$

(JEE MAIN-2020)

Solution

First order decay

$N ( t )= N _{0} e ^{-\lambda 1}$

Given $N ( t ) / N _{0}=9 / 16= e ^{-\lambda}$

Now, $N ( t / 2)= N _{0} e ^{-2} v / 2$

$\frac{ N ( t / 2)}{ N _{0}}=\sqrt{ e ^{-\lambda t}}=\sqrt{9 / 16}$

$N(t/2)=3/4 N _{0}$

Standard 12

Physics

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